In: Statistics and Probability
Walmart Company surveyed the senior managers in the retail industry
to understand their mean annual salary. A random sample of 49
managers reviled a sample mean of $45420. The standard deviation of
this population is $2050. Calculate the 95% confidence interval
estimate.
Solution :
Given that,
= 45420
= 2050
n = 49
At 95% confidence level the z is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
Z/2 = Z0.025 = 1.96
Margin of error = E = Z/2* ( /n)
= 1.96 * (2050 / 49)
= 574
At 95% confidence interval estimate of the population mean is,
- E < < + E
45420 - 574 < < 45420 + 574
44846 < < 45994
($44846 , $45994)